Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main{
  public static void main(String[] args) throws IOException{
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    int n = Integer.parseInt(br.readLine());
    double[] arr = new double[n];
    double ans = 0;
    String[] s = br.readLine().split(" ");
    for(int i=0;i<n;i++)
      arr[i] = Double.parseDouble(s[i]);
    double[][] dp = new double[n+1][n+1];
    
    dp[0][0] = 1.0;
    for(int i=1;i<=n;i++){
      for(int j=0;j<=n;j++){
        if(j==0)
          dp[i][j] = dp[i-1][j]*(1.0-arr[i-1]);
        else
          dp[i][j] = arr[i-1]*dp[i-1][j-1] + (1-arr[i-1])*dp[i-1][j];
      }
    }
    
    for(int i=n/2+1;i<=n;i++)
      ans += dp[n][i];
    System.out.println(ans);
  }
}